Ramanujan and Hardy played a game where they both picked a complex number. If the product of their numbers was $32-8i$, and Hardy picked $5+3i$, what number did Ramanujan pick?
Let Hardy's number be $h$ and Ramanujan's be $r$. We have the equations: \begin{align*}
rh&=32-8i,\\
h&=5+3i.
\end{align*} Thus, \[r=\frac{32-8i}{5+3i}.\] Multiplying top and bottom by the conjugate of $5+3i$, we have \[r=\frac{(32-8i)(5-3i)}{34}\] or  \[r=\frac{136-136i}{34}=\boxed{4-4i}\]